Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $q = \dfrac{8}{15y - 5} \times \dfrac{15y^2 - 5y}{-10} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $q = \dfrac{ 8 \times (15y^2 - 5y) } { (15y - 5) \times -10 } $ $ q = \dfrac {8 \times 5y(3y - 1)} {-10 \times 5(3y - 1)} $ $ q = \dfrac{40y(3y - 1)}{-50(3y - 1)} $ We can cancel the $3y - 1$ so long as $3y - 1 \neq 0$ Therefore $y \neq \dfrac{1}{3}$ $q = \dfrac{40y \cancel{(3y - 1})}{-50 \cancel{(3y - 1)}} = -\dfrac{40y}{50} = -\dfrac{4y}{5} $